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# Algebra

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Let a, b, and c be nonnegative real numbers.  Prove that for all x >= 0,

(x + abc)^3 >= abc(x + a)(x + b)(x + c) >= (x + a + b + c)^3.

Mar 7, 2023

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First, we will prove the left-hand inequality:

(x + abc)^3 >= abc(x + a)(x + b)(x + c)

Expanding both sides of the inequality, we get:

x^3 + 3x^2(abc) + 3x(abc)^2 + (abc)^3 >= abc(x^3 + (a+b+c)x^2 + (ab+bc+ac)x + abc)

Simplifying, we get:

x^3(1-abc) + x^2(3abc + a^2bc + ab^2c + abc^2) + x(3a^2b^2c + 3ab^2c^2 + 3a^2bc^2) + abc^3 >= 0

Since a, b, and c are nonnegative, all the coefficients in this polynomial are nonnegative. Therefore, the polynomial is nonnegative for all x >= 0, which proves the left-hand inequality.

Next, we will prove the right-hand inequality:

abc(x + a)(x + b)(x + c) >= (x + a + b + c)^3

Expanding both sides of the inequality, we get:

abc(x^3 + (a+b+c)x^2 + (ab+bc+ac)x + abc) >= x^3 + 3x^2(a+b+c) + 3x(ab+bc+ac) + (a+b+c)^3

Simplifying, we get:

x^3(1-abc) + x^2(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-3abc) + x(abc(a+b+c)-(a+b+c)^2) + (a+b+c)^3 >= 0

Since a, b, and c are nonnegative, we have:

ab^2+ac^2+ba^2+bc^2+ca^2+cb^2 >= 6abc

and

abc(a+b+c) >= 3abc(x+y+z)

where x = a+b, y = b+c, and z = c+a. Therefore, we can rewrite the inequality as:

x^3(1-abc) + x^2(6abc-3abc) + x(3abc-3abc) + (a+b+c)^3 >= 0

Simplifying, we get:

x^3(1-abc) + 3(a+b+c)^3 >= 0

Since x >= 0 and a, b, and c are nonnegative, both terms in this expression are nonnegative. Therefore, the inequality is true, which proves the right-hand inequality.

Therefore, we have proven both inequalities:

(x + abc)^3 >= abc(x + a)(x + b)(x + c) >= (x + a + b + c)^3

for all nonnegative real numbers a, b, and c and all x >= 0.

Mar 7, 2023